# How to simplify (2x - 3)^2 - 2x(2x - 5)

## Homework Statement

simplify (2x-3)² - 2x(2x-5).

## The Attempt at a Solution

(2x-3)² - 2x(2x-5)
to
2x² + 9 - 4x² + 10x
to
2x² + 9 + 10x

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tiny-tim
Homework Helper
hi tehmatriks!

sorry, but your (2x-3)² is completely wrong

if you can't do it in your head, write it as (2x-3)(2x-3) first, and then expand it

hi tehmatriks!

sorry, but your (2x-3)² is completely wrong

if you can't do it in your head, write it as (2x-3)(2x-3) first, and then expand it
thanks, the whole time i was staring at the (2x-3)² and was just thinking about how wrong i was doing it, i always forget that double bracket thing when it comes to these situations, just dont get these much

thanks again man, here's the final result

(2x-3)² - 2x(2x-5)
to
(2x-3)(2x-3) - 2x(2x-5)
to
4x - 6x - 6x + 9 - 4x + 10x
to
9 - 2x

Mark44
Mentor

thanks, the whole time i was staring at the (2x-3)² and was just thinking about how wrong i was doing it, i always forget that double bracket thing when it comes to these situations, just dont get these much

thanks again man, here's the final result

(2x-3)² - 2x(2x-5)
to
(2x-3)(2x-3) - 2x(2x-5)
to
4x - 6x - 6x + 9 - 4x + 10x
to
9 - 2x
Use = between expressions that have the same value.

(2x-3)2 - 2x(2x-5)
= (2x-3)(2x-3) - 2x(2x-5) -- so far, so good
After that, things go downhill.
(2x)(2x) = 2*2*x*x = ?

And -2x(2x - 5) = -2x * 2x -2x * (-5) = ?
You have to distribute the -2x over both terms inside the parentheses.

Use = between expressions that have the same value.

(2x-3)2 - 2x(2x-5)
= (2x-3)(2x-3) - 2x(2x-5) -- so far, so good
After that, things go downhill.
(2x)(2x) = 2*2*x*x = ?

And -2x(2x - 5) = -2x * 2x -2x * (-5) = ?
You have to distribute the -2x over both terms inside the parentheses.
dammit, it's always the simplest of things that i forget

(2x-3)(2x-3) - 2x(2x-5)
to
4x2 - 6x - 6x + 9 - 4x2 + 10x
to -12x + 10x + 9 = 9 - 2x

thanks mark, you're a legend

Mark44
Mentor

Much better.

The things that you're forgetting are the result of not learning them very well in the first place. If you want to get better at this, I would advise going back and reviewing binomial multiplication (problems like expanding (2x + 3)2) and the distributive law (a(b + c) = ab + ac).